Using jQuery to store basic text string in mySQL base?
- by Kenny Bones
Could someone point me in the right direction here?
Basically, I've got this jQuery code snippet:
$('.bggallery_images').click(function () {
var newBG = "url('" + $(this).attr('src');
var fullpath = $(this).attr('src');
var filename = fullpath.replace('img/Bakgrunner/', '');
$('#wrapper').css('background-image', newBG);
// Lagre til SQL
$.ajax({
url: "save_to_db.php",
// The url to your function to handle saving to the db
data: filename,
dataType: 'Text',
type: 'POST',
// Could also use GET if you prefer
success: function (data) {
// Just for testing purposes.
alert('Background changed to: ' + data);
}
});
});
This is being run when I click a certain button. So it's actually within a click handler.
If I understand this correctly, this snippet takes the source if the image I just clicked and strips it so I end up with only the filename. If I do an alert(filename), I get the filename only. So this is working ok.
But then, it does an ajax call to a php file called "save_to_db.php" and sends data: filename. This is correct right? Then, it does a callback which does an alert + data.
Does this seem correct so far?
Cause my php file looks like this:
<?php
require("dbconnect2.php");
$uploadstring = $_POST['filename'];
$sessionid = $_SESSION['id'];
echo ($sessionid);
mysql_query("UPDATE brukere SET brukerBakgrunn = '$uploadstring' WHERE brukerID=" .$_SESSION['id']);
mysql_close();
?>
When I click the image, the jQuery snippet fires and I get the results of this php file as output for the alert box. I think that the variables somehow are empty.
Because notice the echo($sessionid); which is a variable I've created just to test what the session ID is. And it returns nothing. What could be the issue here?
Edit: I just tried to echo out the $uploadstring variable as well and it also returns nothing. It's like the jQuery snippet doesn't even pass the variable on to the php file?
